Image of a discrete group in a vector space is discrete and closed

Question

Let $\Gamma \subset \operatorname{GL}_n(\mathbb R)$ be a discrete subgroup, and let $0 \neq v \in \mathbb R^n$. Is it true that $\Gamma.v = \{ \gamma.v : \gamma \in \Gamma\}$ is discrete and closed in $\mathbb R^n$?

Writing out $v = (v_1, ... , v_n)$, and $\gamma = (\gamma_{ij})$, it suffices to show that for each $j$, the elements $\sum\limits_{i=1}^n \gamma_{ij}v_i$ run through a discrete and closed set in $\mathbb R$.

While each of the sets $\{ \gamma_{ij}v_i\} \subset \mathbb R$ is discrete, as $\gamma$ runs through the elements of $\Gamma$, it is generally not true that $E+J$ is discrete if $E$ and $J$ are discrete subsets of $\mathbb R$.

Answer 1

No, not at all, even for cyclic groups. As a simple example, take a linear transformation $A\in GL(3, {\mathbb R})$ which has one eigenvalue $>1$ and two complex conjugate eigenvalues with absolute value $1$, and which are not roots of unity. Let $\Gamma$ be the subgroup of $GL(3, {\mathbb R})$ generated by $A$. I will leave it to you to prove the discreteness of $\Gamma$ and to find a (nonzero) vector whose $\Gamma$-orbit is dense in an ellipse.

As a more interesting example, take $\Gamma= SL(2, {\mathbb Z})$. One can prove that for almost every vector $v\in {\mathbb R}^2$, its $\Gamma$-orbit is dense in ${\mathbb R}^2$ (because the $\Gamma$-action is ergodic).

Answer 2

For another simple example, let $n=1$, and $\Gamma$ be the subgroup of $\text{GL}_1(\mathbb R) = \mathbb R\setminus \{0\}$, generated by any $g>1$. Then $\Gamma$ is discrete but, for any nonzero $x$, the orbit $$ \{g^nx:n\in \mathbb Z\} $$ has zero in its closure.